A Journey Towards One

If you analyze, you’ll find it is better to divide $N$ by 2 when available.

So,

dp[n] = 1 + dp[n / 2], when N is even.

When $N$ is odd, we have to increase/decrease N by 1 and then divide it by 2.

dp[n] = 2 + min(dp[(n - 1)/2], dp[(n + 1)/2])

This can be solved easily with a recursive solution with the base case, if $N = 1$, then the answer is 0.